Class Solution


  • public class Solution
    extends Object
    2483 - Minimum Penalty for a Shop.

    Medium

    You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y':

    • if the ith character is 'Y', it means that customers come at the ith hour
    • whereas 'N' indicates that no customers come at the ith hour.

    If the shop closes at the jth hour (0 <= j <= n), the penalty is calculated as follows:

    • For every hour when the shop is open and no customers come, the penalty increases by 1.
    • For every hour when the shop is closed and customers come, the penalty increases by 1.

    Return the earliest hour at which the shop must be closed to incur a minimum penalty.

    Note that if a shop closes at the jth hour, it means the shop is closed at the hour j.

    Example 1:

    Input: customers = “YYNY”

    Output: 2

    Explanation:

    • Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty.
    • Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty.
    • Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty.
    • Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty.
    • Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty.

    Closing the shop at 2nd or 4th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.

    Example 2:

    Input: customers = “NNNNN”

    Output: 0

    Explanation: It is best to close the shop at the 0th hour as no customers arrive.

    Example 3:

    Input: customers = “YYYY”

    Output: 4

    Explanation: It is best to close the shop at the 4th hour as customers arrive at each hour.

    Constraints:

    • 1 <= customers.length <= 105
    • customers consists only of characters 'Y' and 'N'.
    • Constructor Detail

      • Solution

        public Solution()
    • Method Detail

      • bestClosingTime

        public int bestClosingTime​(String customers)